Ellipse 2 Question 8
8. The eccentricity of an ellipse whose centre is at the origin is $1 / 2$. If one of its directrices is $x=-4$, then the equation of the normal to it at $1, \frac{3}{2}$ is
(a) $2 y-x=2$
(b) $4 x-2 y=1$
(c) $4 x+2 y=7$
(d) $x+2 y=4$
(2017 Main)
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Solution:
- We have, $e=\frac{1}{2}$ and $\frac{a}{e}=4$
$\therefore \quad a=2$
Now, $b^{2}=a^{2}\left(1-e^{2}\right)=(2)^{2} 1-\frac{1}{2}^{2}=41-\frac{1}{4}=3$
$\Rightarrow \quad b=\sqrt{3}$
$\therefore$ Equation of the ellipse is $\frac{x^{2}}{(2)^{2}}+\frac{y^{2}}{(\sqrt{3})^{2}}=1$
$$ \Rightarrow \quad \frac{x^{2}}{4}+\frac{y^{2}}{3}=1 $$
Now, the equation of normal at $1, \frac{3}{2}$ is
$$ \begin{array}{rlrl} & & \frac{a^{2} x}{x _1}-\frac{b^{2} y}{y _1} & =a^{2}-b^{2} \\ \Rightarrow & & \frac{4 x}{1}-\frac{3 y}{(3 / 2)} & =4-3 \\ \Rightarrow & 4 x-2 y & =1 \end{array} $$
